How to make something that's equivalent to the and operator but without using and itself?

Question

maybe a function, but to also respect the lazy eval of and. So a and b vs fand(a,b) or fand([a,b]) but get lazy eval too? Is it possible? Without lazy eval is easy, a one line function.

Answer 1

# If you want to use ’and’ to open up (evaluate) the second argument (short circut), you can use an if-else statement, as you already mentioned:

>>> a, b = 3, 4
>>> print(a and b)
4
>>> print(a if not a else b)
4
>>> print(b if a else a)
4
>>> print(b if all((a, b)) else 0)
4
>>> print((lambda a, b: b if all((a, b)) else 0)(a, b))
4

Answer 2

Now I see you want to accept and work with any type, not bool in particular.

Good luck trying 👍

Answer 3

Every value in Python is truthly or falsy. Truthly values can be evaluated to True and falsy values to False. For integers, the only falsy value is 0.

Answer 4

Oh that __and__ overload didn't work because it was overriding & (the bitwise and), instead of logical 'and'. Great...

Answer 5

NewlyRefocused,
Sorry I was mistaken, it apparently is supposed to be for the bitwise AND operator overloading.

I just found these recently,

https://stackoverflow.com/questions/471546/any-way-to-override-the-and-operator-in-JUMP_LINK__&&__python__&&__JUMP_LINK
https://stackoverflow.com/questions/32311518/is-it-possible-to-overload-logical-and-in-python

Answer 6

Ipang that Infix thing is amazing, now all I've to do is understand it then see how I could do it here.

Answer 7

Ipang it looks nice to use like "50 |infix_and| 0"(ignore the dbl.quotes) but I still don't have the lazy eval. I'll try to understand it. (Infix stuff is at the end of this code blob):
https://code.sololearn.com/cPZGkB6yyETc/?ref=app

Answer 8

You mean __and__ magic overload is not an option, or it is, but there should be no `and` in its implementation?

Answer 9

# Hi! One way is to use all():

print(all((True, True)) == (True and True))
print(all((True, False)) == (True and False))
print(all((False, True)) == (False and True))
print(all((False, False)) == (False and False))

Answer 10

NewlyRefocused,
Why return 9 at line 9?

Also <value> argument in __init()__ needs to be casted as `bool` to ensure we only work with logical values.

Answer 11

the return 9 is to see if that function is called, so the print should show 9 instead of [] Also remember that '7 and 8' returns 8 not True, hence why not forcing bool casting

Answer 12

Yes but, how to fix the overload? it's not using the overload as you'd expect

Answer 13

That didn't pass tests, eg.:
A=7
B=8
assert (A and B) == (not(not A or not B))

Answer 14

a=5
b=4
print((a!=0)*b) #same as (a and b)

Output:
4

Another one:
print(bool(a)*b)

Answer 15

I'm interested in the "and" operator, instead of the "&" operator. Specifically this is how it works: "assert (a and b) == (b if a else a)" for any value or type of a,b I hope.

Answer 16

Okay bro
Let me know how it goes I wanna know also ...

Answer 17

You could use these asserts to test I guess: https://code.sololearn.com/cPZGkB6yyETc/?ref=app

Answer 18

I don't know what magic overload is. all() returns bool so it's no good eg 3 and 4 is 4 not True like all([3,4]) would be.

Answer 19

yeah, that's pretty good like that, but I was hoping to compress it into one word so as to not repeat the args (in this case 'a') But I guess 'a fand b' cant be done in python?

Answer 20

Someone said that maybe generators could be used for the lazy eval. I've not reached that far in my learning, does anyone know how and if that could work here?